If the constant term in the expansion of [([5]3x+2 x[3]5)^12, x 0], is [ 2^8 [5]3], then […

Question

If the constant term in the expansion of $$\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}, x \neq 0$$, is $$\alpha \times 2^8 \times \sqrt[5]{3}$$, then $$25 \alpha$$ is equal to :

Accepted Answer

$$\begin{aligned}
& \left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[5]{3}}\right)^{12} \
& T_{r+1}={ }^{12} C r\left(\frac{\sqrt[5]{3}}{x}\right)^{12-r}\left(\frac{2 x}{\sqrt[3]{5}}\right)^r
\end{aligned}$$

For constant term $-12+r+r=0$

$$\begin{aligned}
& \Rightarrow \quad r=6 \
& \therefore \quad \text { Constant term }={ }^{12} C_6 \frac{(3)^{\frac{6}{5}}}{(5)^{\frac{6}{3}}}(2)^6
\end{aligned}$$

$$\begin{aligned}
& ={ }^{12} C_6 \times \frac{2^6}{25} \times 3.3^{\frac{1}{5}} \
& =\frac{231}{25} \times 2^8 \cdot 3^{\frac{1}{5}} \cdot 3 \
& =\frac{693}{25} \cdot 2^8 \sqrt[5]{3} \
& \therefore \quad \alpha=\frac{693}{25} \
& 25 \alpha=693
\end{aligned}$$

Solutions · 1